Samacheer Kalvi Class 12 Maths Solutions Chapter 11 Probability Distributions Exercise 11.1

Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.1

 

Question 1. Suppose X is the number of tails that occurred when three fair coins are tossed once simultaneously. Find the values of the random variable X and number of points in it. inverse-images.
Answer: Let X be the random variable that shows the number of tails when three coins are tossed at the same time.
The sample space S consists of all possible outcomes: \( S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\} \).
The random variable X takes values based on the number of tails in each outcome:
\( X(HHH) = 0 \) (zero tails)
\( X(HHT) = 1, X(HTH) = 1, X(THH) = 1 \) (one tail)
\( X(HTT) = 2, X(THT) = 2, X(TTH) = 2 \) (two tails)
\( X(TTT) = 3 \) (three tails)
So, X can take the values 0, 1, 2, 3. The sum of the number of elements in the inverse image is the total number of outcomes, which is 8.
The table below shows the values of the random variable X and the number of points in its inverse images:

In simple words: We count how many tails appear when three coins are tossed. The random variable X lists these counts (0, 1, 2, or 3 tails), and we see how many ways each count can happen.

🎯 Exam Tip: When defining a random variable, always clearly state what it represents and list all possible values it can take based on the experiment.

 

Question 2. In a pack of 52 playing cards, two cards are drawn at random simultaneously. If the number of black cards drawn is a random variable, find the values of the random variable and number of points in its inverse images.
Answer: Let X be the random variable that represents the number of black cards drawn.
A standard deck has 52 cards: 26 black cards and 26 red cards.
When two cards are drawn, the number of black cards (X) can be 0, 1, or 2. So, \( X = \{0, 1, 2\} \).
The total number of ways to draw 2 cards from 52 is given by \( {}^{52}C_2 \).
\( {}^{52}C_2 = \frac{52 \times 51}{2 \times 1} = 26 \times 51 = 1326 \).
Now, let's find the number of points for each value of X:
For \( X = 0 \): This means both cards drawn are white (red cards).
Number of ways = \( {}^{26}C_2 = \frac{26 \times 25}{2 \times 1} = 13 \times 25 = 325 \).
For \( X = 1 \): This means one card is black and one card is white (red).
Number of ways = \( {}^{26}C_1 \times {}^{26}C_1 = 26 \times 26 = 676 \).
For \( X = 2 \): This means both cards drawn are black.
Number of ways = \( {}^{26}C_2 = \frac{26 \times 25}{2 \times 1} = 13 \times 25 = 325 \).
The table below shows the values of the random variable X and the number of points in its inverse images:

In simple words: When you pick two cards from a deck, the random variable X tells you how many of them are black. We find out how many ways you can get 0, 1, or 2 black cards.

🎯 Exam Tip: Remember to use combination formulas (\( {}^{n}C_r \)) correctly for "choosing" items without regard to order, and multiply combinations when choices are made from different groups.

 

Question 3. An urn contains 5 mangoes and 4 apples. Three fruits are taken at random. If the number of apples taken is a random variable, then find the values of the random variable and number of points in its inverse images.
Answer: Number of mangoes = 5
Number of apples = 4
Total number of fruits = \( 5 + 4 = 9 \).
Let X be the random variable that denotes the number of apples taken when three fruits are chosen.
X can take values 0, 1, 2, or 3, as we are picking 3 fruits and there are 4 apples available.
Let M represent a mango and A represent an apple.
For \( X = 0 \): This means all three fruits taken are mangoes (MMM).
Number of ways = \( {}^{5}C_3 \times {}^{4}C_0 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} \times 1 = 10 \times 1 = 10 \).
For \( X = 1 \): This means one apple and two mangoes are taken (AMM, MAM, MMA).
Number of ways = \( {}^{4}C_1 \times {}^{5}C_2 = 4 \times \frac{5 \times 4}{2 \times 1} = 4 \times 10 = 40 \).
For \( X = 2 \): This means two apples and one mango are taken (AAM, AMA, MAA).
Number of ways = \( {}^{4}C_2 \times {}^{5}C_1 = \frac{4 \times 3}{2 \times 1} \times 5 = 6 \times 5 = 30 \).
For \( X = 3 \): This means all three fruits taken are apples (AAA).
Number of ways = \( {}^{4}C_3 \times {}^{5}C_0 = 4 \times 1 = 4 \).
The total number of ways to pick 3 fruits from 9 is \( {}^{9}C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84 \). This matches the sum of the counts above \( 10+40+30+4 = 84 \).
The table below shows the values of the random variable X and the number of points in its inverse images:

In simple words: From a bag of mangoes and apples, we take three fruits. The random variable X counts how many of those three are apples. We calculate how many ways we can get 0, 1, 2, or 3 apples.

🎯 Exam Tip: When dealing with selections from two different categories, multiply the combinations for each category to find the total ways for that specific outcome.

 

Question 4. Two balls are chosen randomly from an urn containing 6 red and 8 black balls. Suppose that we win Rs 15 for each red ball selected and we lose Rs 10 for each black ball selected. If X denotes the winning amount, then find the values of X and the number of points in its inverse images.
Answer: Number of red balls = 6
Number of black balls = 8
Total number of balls = \( 6 + 8 = 14 \).
Let X be the random variable representing the winning amount. We are choosing 2 balls.
Possible outcomes and corresponding winning amounts:
1. Both balls are black:
Winning amount = \( 2 \times (-\text{Rs } 10) = -\text{Rs } 20 \). (This means a loss of Rs 20)
2. One red ball and one black ball:
Winning amount = \( \text{Rs } 15 + (-\text{Rs } 10) = \text{Rs } 5 \).
3. Both balls are red:
Winning amount = \( 2 \times \text{Rs } 15 = \text{Rs } 30 \).
So, the possible values for the random variable X are \( \{-20, 5, 30\} \).
Now, let's find the number of ways for each outcome:
The total number of ways to choose 2 balls from 14 is \( {}^{14}C_2 \).
\( {}^{14}C_2 = \frac{14 \times 13}{2 \times 1} = 7 \times 13 = 91 \).
For \( X = -20 \) (both black balls):
Number of ways = \( {}^{8}C_2 = \frac{8 \times 7}{2 \times 1} = 4 \times 7 = 28 \).
For \( X = 5 \) (one red and one black ball):
Number of ways = \( {}^{6}C_1 \times {}^{8}C_1 = 6 \times 8 = 48 \).
For \( X = 30 \) (both red balls):
Number of ways = \( {}^{6}C_2 = \frac{6 \times 5}{2 \times 1} = 3 \times 5 = 15 \).
The sum of the ways is \( 28 + 48 + 15 = 91 \), which matches the total possible ways.
The table below shows the values of the random variable X and the number of points in its inverse images:

In simple words: We pick two balls from a mix of red and black balls. We win money for red balls and lose for black ones. The random variable X tells us the total money we win or lose. We then find how many ways each winning or losing amount can occur.

🎯 Exam Tip: Be careful with the signs for winning and losing amounts; losses should be represented by negative values. Always check if the sum of inverse image counts matches the total sample space size.

 

Question 5. A six-sided die is marked '2' on one face, '3' on two of its faces, and '4' on the remaining three faces. The die is thrown twice. If X denotes the total score in two throws, find the values of the random variable and number of points in its inverse images.
Answer: Let X be the random variable denoting the total score in two throws of the special die.
The faces of the die are marked as follows:
- One face has '2'.
- Two faces have '3'.
- Three faces have '4'.
Total number of faces = \( 1 + 2 + 3 = 6 \).
When the die is thrown twice, the sample space S consists of pairs of scores. The total number of possible outcomes is \( 6 \times 6 = 36 \).
Let's create a table for the sample space showing the sum (total score) for each throw combination:

From the sample space table, the distinct values for X (total score) are: \( \{4, 5, 6, 7, 8\} \).
Now, we count the number of times each score appears (inverse images):
For \( X = 4 \): There is 1 occurrence (2+2).
For \( X = 5 \): There are 4 occurrences (2+3, 2+3, 3+2, 3+2).
For \( X = 6 \): There are 10 occurrences (2+4, 2+4, 2+4, 3+3, 3+3, 3+3, 3+3, 4+2, 4+2, 4+2).
For \( X = 7 \): There are 12 occurrences (3+4, 3+4, 3+4, 3+4, 3+4, 3+4, 4+3, 4+3, 4+3, 4+3, 4+3, 4+3).
For \( X = 8 \): There are 9 occurrences (4+4, 4+4, 4+4, 4+4, 4+4, 4+4, 4+4, 4+4, 4+4).
Total number of elements in the sample space \( n(S) = 36 \).
The table below shows the values of the random variable X and the number of points in its inverse images:In simple words: A special die is rolled two times, and we add up the numbers that show on top. The random variable X lists all the possible sums we can get. We count how many ways each sum can happen.

🎯 Exam Tip: When dealing with weighted dice or multiple trials, creating a full sample space table helps ensure all combinations and their sums are correctly identified and counted.